Chapter 13.7 - Theorem 2

In the previous section, we saw examples on limits of polynomial functions and rational functions. In this section, we will see the second theorem.

Theorem 2

For any positive integer n,

$$\lim_{x\rightarrow a} \left[\frac{x^n - a^n}{x-a} \right]~=~n a^{n-1}$$

• The expression in this theorem is true even if n is any rational number and a is positive.

• We will see the proof for this theorem in higher classes. At present we will see some solved examples.

Solved example 13.3
(i) Evaluate $\lim_{x\rightarrow a} \left[\frac{x^{15} - 1}{x^{10} - 1} \right]$
(ii) Evaluate $\lim_{x\rightarrow 0} \left[\frac{\sqrt{1+x} - 1}{x} \right]$ 

Solution:
Part (i):
We have:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 1} \left[\frac{x^{15} - 1}{x^{10} - 1} \right]}
& {~=~}& {\lim_{x\rightarrow 1} \left[\frac{x^{15} - 1}{x - 1}~\div~\frac{x^{10} - 1}{x - 1} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 1} \left[\frac{x^{15} - 1}{x - 1} \right]~\div~\lim_{x\rightarrow 1} \left[\frac{x^{10} - 1}{x - 1} \right]}
&{} \\

{}&{}
& {~=~}& {\left[15 × 1^{15-1}\right]~\div~\left[10 × 1^{10-1}\right]}
&{} \\

{}&{}
& {~=~}& {\left[15 × 1^{14}\right]~\div~\left[10 × 1^{9}\right]}
&{} \\

{}&{}
& {~=~}& {\left[15 × 1\right]~\div~\left[10 × 1\right]}
&{} \\

{}&{}
& {~=~}& {\frac{15}{10}}
&{} \\

{}&{}
& {~=~}& {\frac{3}{2}}
&{} \\

\end{array}$

Part (ii):
1. Let u = x +1
2. We want the limit when x approaches zero.
• Based on (1), we can write:
When x approaches zero, u will approach 1
• Also based on (1), we get: x = u-1
3. So the required limit can be written in another form:
$\lim_{u\rightarrow 1} \left[\frac{\sqrt{u} - 1}{u-1} \right]$
4. Now we can evaluate the limit:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 0} \left[\frac{\sqrt{1+x} - 1}{x} \right]}
& {~=~}& {\lim_{u\rightarrow 1} \left[\frac{\sqrt{u} - 1}{u-1} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{u\rightarrow 1} \left[\frac{u^{\frac{1}{2}} - 1}{u-1} \right]}
&{} \\

{}&{}
& {~=~}& {\frac{1}{2} × 1^{\left(\frac{1}{2} - 1\right)}}
&{} \\

{}&{}
& {~=~}& {\frac{1}{2} × 1^{\left(-\frac{1}{2}\right)}}
&{} \\

{}&{}
& {~=~}& {\frac{1}{2} × 1}
&{} \\

{}&{}
& {~=~}& {\frac{1}{2}}
&{} \\

\end{array}$

Alternate method:
$\begin{array}{ll}
{}&{\lim_{x\rightarrow 0} \left[\frac{\sqrt{1+x} - 1}{x} \right]}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{\sqrt{1+x} - 1}{x} \times \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1}  \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{1+x - 1}{x(\sqrt{1+x} + 1)} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{x}{x(\sqrt{1+x} + 1)} \right]}
&{} \\

{}&{}
& {~=~}& {\lim_{x\rightarrow 0} \left[\frac{1}{(\sqrt{1+x} + 1)} \right]}
&{} \\


{}&{}
& {~=~}& {\left[\frac{1}{(\sqrt{1+0} + 1)} \right]}
&{} \\

{}&{}
& {~=~}& {\left[\frac{1}{(1 + 1)} \right]}
&{} \\

{}&{}
& {~=~}& {\frac{1}{2}}
&{} \\

\end{array}$

◼ Remarks:
• In the first line, we multiply both numerator and denominator by "√(1+x) + 1".
    ♦ So the numerator will be in the form (a+b)(a-b)
    ♦ We know that, (a+b)(a-b) = a² - b².

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In the next section, we will see limits of trigonometric functions. 

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